How to decompose a MIMO system into a multiplication minimum and nonminimum phase part
$begingroup$
suppose the transfer function of m-input-m-output system is
$$Gleft( s right) = left[ {begin{array}{*{20}{c}}
{{g_{11}}left( s right)}& cdots &{{g_{1m}}left( s right)}\
vdots & ddots & vdots \
{{g_{m1}}left( s right)}& cdots &{{g_{mm}}left( s right)}
end{array}} right]$$
Each ${g_{ij}}(s)$ might contains nonminimum phase part.Then, how could the system be decomposed into the multiplication of minimum ${G_{min }}left( s right)$ and nonminimum phase part ${G_{nonmin }}left( s right)$
$$Gleft( s right) = {G_{min }}left( s right) * {G_{nonmin }}left( s right)$$
Is there any command in MATLAB to realize it?
matlab control-theory
asked Jan 18 at 12:00
user4756766user4756766
357
$endgroup$
add a comment |
$begingroup$
suppose the transfer function of m-input-m-output system is
$$Gleft( s right) = left[ {begin{array}{*{20}{c}}
{{g_{11}}left( s right)}& cdots &{{g_{1m}}left( s right)}\
vdots & ddots & vdots \
{{g_{m1}}left( s right)}& cdots &{{g_{mm}}left( s right)}
end{array}} right]$$
Each ${g_{ij}}(s)$ might contains nonminimum phase part.Then, how could the system be decomposed into the multiplication of minimum ${G_{min }}left( s right)$ and nonminimum phase part ${G_{nonmin }}left( s right)$
$$Gleft( s right) = {G_{min }}left( s right) * {G_{nonmin }}left( s right)$$
Is there any command in MATLAB to realize it?
matlab control-theory
asked Jan 18 at 12:00
user4756766user4756766
357
$endgroup$
$begingroup$
There is no unique way of doing this, it will only tell you how to split the zeros, but not the poles.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:09
$begingroup$
If $m=1$, then it is easy to decompose by simply finding the unstable zeros. However, for MIMO system, even if I can find unstable zeros for each $g_{ij}(s)$, it's still hard to obtain the final multiplication form.
$endgroup$
– user4756766
Jan 18 at 12:14
$begingroup$
So you do not care if one of the final transfer functions is not causal, otherwise you still also have to choose how to split up the poles, even for $m=1$.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:16
$begingroup$
What if I do not care about whether it's causal. Is there any simple method to realize the decomposition? Could you give me some hint about that @KwinvanderVeen
$endgroup$
– user4756766
Jan 18 at 12:22
add a comment |
$begingroup$
suppose the transfer function of m-input-m-output system is
$$Gleft( s right) = left[ {begin{array}{*{20}{c}}
{{g_{11}}left( s right)}& cdots &{{g_{1m}}left( s right)}\
vdots & ddots & vdots \
{{g_{m1}}left( s right)}& cdots &{{g_{mm}}left( s right)}
end{array}} right]$$
Each ${g_{ij}}(s)$ might contains nonminimum phase part.Then, how could the system be decomposed into the multiplication of minimum ${G_{min }}left( s right)$ and nonminimum phase part ${G_{nonmin }}left( s right)$
$$Gleft( s right) = {G_{min }}left( s right) * {G_{nonmin }}left( s right)$$
Is there any command in MATLAB to realize it?
matlab control-theory
asked Jan 18 at 12:00
user4756766user4756766
357
$endgroup$
suppose the transfer function of m-input-m-output system is
$$Gleft( s right) = left[ {begin{array}{*{20}{c}}
{{g_{11}}left( s right)}& cdots &{{g_{1m}}left( s right)}\
vdots & ddots & vdots \
{{g_{m1}}left( s right)}& cdots &{{g_{mm}}left( s right)}
end{array}} right]$$
Each ${g_{ij}}(s)$ might contains nonminimum phase part.Then, how could the system be decomposed into the multiplication of minimum ${G_{min }}left( s right)$ and nonminimum phase part ${G_{nonmin }}left( s right)$
$$Gleft( s right) = {G_{min }}left( s right) * {G_{nonmin }}left( s right)$$
Is there any command in MATLAB to realize it?
matlab control-theory
matlab control-theory
asked Jan 18 at 12:00
user4756766user4756766
357
asked Jan 18 at 12:00
user4756766user4756766
357
asked Jan 18 at 12:00
user4756766user4756766
357
asked Jan 18 at 12:00
user4756766user4756766
357
asked Jan 18 at 12:00
user4756766user4756766
357
357
$begingroup$
There is no unique way of doing this, it will only tell you how to split the zeros, but not the poles.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:09
$begingroup$
If $m=1$, then it is easy to decompose by simply finding the unstable zeros. However, for MIMO system, even if I can find unstable zeros for each $g_{ij}(s)$, it's still hard to obtain the final multiplication form.
$endgroup$
– user4756766
Jan 18 at 12:14
$begingroup$
So you do not care if one of the final transfer functions is not causal, otherwise you still also have to choose how to split up the poles, even for $m=1$.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:16
$begingroup$
What if I do not care about whether it's causal. Is there any simple method to realize the decomposition? Could you give me some hint about that @KwinvanderVeen
$endgroup$
– user4756766
Jan 18 at 12:22
add a comment |
$begingroup$
There is no unique way of doing this, it will only tell you how to split the zeros, but not the poles.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:09
$begingroup$
If $m=1$, then it is easy to decompose by simply finding the unstable zeros. However, for MIMO system, even if I can find unstable zeros for each $g_{ij}(s)$, it's still hard to obtain the final multiplication form.
$endgroup$
– user4756766
Jan 18 at 12:14
$begingroup$
So you do not care if one of the final transfer functions is not causal, otherwise you still also have to choose how to split up the poles, even for $m=1$.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:16
$begingroup$
What if I do not care about whether it's causal. Is there any simple method to realize the decomposition? Could you give me some hint about that @KwinvanderVeen
$endgroup$
– user4756766
Jan 18 at 12:22
$begingroup$
There is no unique way of doing this, it will only tell you how to split the zeros, but not the poles.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:09
$begingroup$
There is no unique way of doing this, it will only tell you how to split the zeros, but not the poles.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:09
$begingroup$
If $m=1$, then it is easy to decompose by simply finding the unstable zeros. However, for MIMO system, even if I can find unstable zeros for each $g_{ij}(s)$, it's still hard to obtain the final multiplication form.
$endgroup$
– user4756766
Jan 18 at 12:14
$begingroup$
If $m=1$, then it is easy to decompose by simply finding the unstable zeros. However, for MIMO system, even if I can find unstable zeros for each $g_{ij}(s)$, it's still hard to obtain the final multiplication form.
$endgroup$
– user4756766
Jan 18 at 12:14
$begingroup$
So you do not care if one of the final transfer functions is not causal, otherwise you still also have to choose how to split up the poles, even for $m=1$.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:16
$begingroup$
So you do not care if one of the final transfer functions is not causal, otherwise you still also have to choose how to split up the poles, even for $m=1$.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:16
$begingroup$
What if I do not care about whether it's causal. Is there any simple method to realize the decomposition? Could you give me some hint about that @KwinvanderVeen
$endgroup$
– user4756766
Jan 18 at 12:22
$begingroup$
What if I do not care about whether it's causal. Is there any simple method to realize the decomposition? Could you give me some hint about that @KwinvanderVeen
$endgroup$
– user4756766
Jan 18 at 12:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you define the zeros of a transfer function as the values for $s$ at which it loses rank (also known as transmission zeros) then you could make use of the Smith McMillan form. Namely by multiplying any MIMO transfer function matrix by its pole polynomial one can obtain a polynomial matrix $P(s)$. Any polynomial can be written as
$$
P(s) = U_1(s),D(s),U_2(s)
$$
where $U_1(s)$ and $U_2(s)$ are unimodular (does not lose rank for any $s$) and $D(s)$ is diagonal (does loses rank at $s$ equal to a zero). This diagonal matrix can be split up into two diagonal matrices, one with only right half plane zeros and the other with only left half plane zeros.
For example
$$
G(s) =
begin{bmatrix}
frac{s^4+s+1}{(s^2+1)(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+3)} \
frac{s-1}{(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+1)(s+3)}
end{bmatrix}
$$
can also be written as
begin{align}
G(s) &= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix} \
&= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & 1
end{bmatrix}
begin{bmatrix}
1 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix}
end{align}
Multiplying the left unimodular matrix with the left diagonal matrix and the right diagonal matrix with right unimodular matrix and distribute the terms of the pole polynomial gives
$$
G(s) =
begin{bmatrix}
frac{s+2}{s+1} & 1 \
0 & frac{1}{s+1}
end{bmatrix}
begin{bmatrix}
frac{1}{(s^2+1)(s+3)} & 0 \
frac{s-1}{s+3} & frac{1-s}{(s^2+1)(s+3)}
end{bmatrix}
$$
Of course you could use a different distribution of the pole polynomial and you could throw any two unimodular matrices, which are each others inverse, in-between the two two diagonal matrices. So splitting a transfer function matrix into a minimum phase and non-minimum phase part will in general not be unique.
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
If you define the zeros of a transfer function as the values for $s$ at which it loses rank (also known as transmission zeros) then you could make use of the Smith McMillan form. Namely by multiplying any MIMO transfer function matrix by its pole polynomial one can obtain a polynomial matrix $P(s)$. Any polynomial can be written as
$$
P(s) = U_1(s),D(s),U_2(s)
$$
where $U_1(s)$ and $U_2(s)$ are unimodular (does not lose rank for any $s$) and $D(s)$ is diagonal (does loses rank at $s$ equal to a zero). This diagonal matrix can be split up into two diagonal matrices, one with only right half plane zeros and the other with only left half plane zeros.
For example
$$
G(s) =
begin{bmatrix}
frac{s^4+s+1}{(s^2+1)(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+3)} \
frac{s-1}{(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+1)(s+3)}
end{bmatrix}
$$
can also be written as
begin{align}
G(s) &= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix} \
&= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & 1
end{bmatrix}
begin{bmatrix}
1 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix}
end{align}
Multiplying the left unimodular matrix with the left diagonal matrix and the right diagonal matrix with right unimodular matrix and distribute the terms of the pole polynomial gives
$$
G(s) =
begin{bmatrix}
frac{s+2}{s+1} & 1 \
0 & frac{1}{s+1}
end{bmatrix}
begin{bmatrix}
frac{1}{(s^2+1)(s+3)} & 0 \
frac{s-1}{s+3} & frac{1-s}{(s^2+1)(s+3)}
end{bmatrix}
$$
Of course you could use a different distribution of the pole polynomial and you could throw any two unimodular matrices, which are each others inverse, in-between the two two diagonal matrices. So splitting a transfer function matrix into a minimum phase and non-minimum phase part will in general not be unique.
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
add a comment |
$begingroup$
If you define the zeros of a transfer function as the values for $s$ at which it loses rank (also known as transmission zeros) then you could make use of the Smith McMillan form. Namely by multiplying any MIMO transfer function matrix by its pole polynomial one can obtain a polynomial matrix $P(s)$. Any polynomial can be written as
$$
P(s) = U_1(s),D(s),U_2(s)
$$
where $U_1(s)$ and $U_2(s)$ are unimodular (does not lose rank for any $s$) and $D(s)$ is diagonal (does loses rank at $s$ equal to a zero). This diagonal matrix can be split up into two diagonal matrices, one with only right half plane zeros and the other with only left half plane zeros.
For example
$$
G(s) =
begin{bmatrix}
frac{s^4+s+1}{(s^2+1)(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+3)} \
frac{s-1}{(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+1)(s+3)}
end{bmatrix}
$$
can also be written as
begin{align}
G(s) &= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix} \
&= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & 1
end{bmatrix}
begin{bmatrix}
1 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix}
end{align}
Multiplying the left unimodular matrix with the left diagonal matrix and the right diagonal matrix with right unimodular matrix and distribute the terms of the pole polynomial gives
$$
G(s) =
begin{bmatrix}
frac{s+2}{s+1} & 1 \
0 & frac{1}{s+1}
end{bmatrix}
begin{bmatrix}
frac{1}{(s^2+1)(s+3)} & 0 \
frac{s-1}{s+3} & frac{1-s}{(s^2+1)(s+3)}
end{bmatrix}
$$
Of course you could use a different distribution of the pole polynomial and you could throw any two unimodular matrices, which are each others inverse, in-between the two two diagonal matrices. So splitting a transfer function matrix into a minimum phase and non-minimum phase part will in general not be unique.
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
add a comment |
$begingroup$
If you define the zeros of a transfer function as the values for $s$ at which it loses rank (also known as transmission zeros) then you could make use of the Smith McMillan form. Namely by multiplying any MIMO transfer function matrix by its pole polynomial one can obtain a polynomial matrix $P(s)$. Any polynomial can be written as
$$
P(s) = U_1(s),D(s),U_2(s)
$$
where $U_1(s)$ and $U_2(s)$ are unimodular (does not lose rank for any $s$) and $D(s)$ is diagonal (does loses rank at $s$ equal to a zero). This diagonal matrix can be split up into two diagonal matrices, one with only right half plane zeros and the other with only left half plane zeros.
For example
$$
G(s) =
begin{bmatrix}
frac{s^4+s+1}{(s^2+1)(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+3)} \
frac{s-1}{(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+1)(s+3)}
end{bmatrix}
$$
can also be written as
begin{align}
G(s) &= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix} \
&= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & 1
end{bmatrix}
begin{bmatrix}
1 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix}
end{align}
Multiplying the left unimodular matrix with the left diagonal matrix and the right diagonal matrix with right unimodular matrix and distribute the terms of the pole polynomial gives
$$
G(s) =
begin{bmatrix}
frac{s+2}{s+1} & 1 \
0 & frac{1}{s+1}
end{bmatrix}
begin{bmatrix}
frac{1}{(s^2+1)(s+3)} & 0 \
frac{s-1}{s+3} & frac{1-s}{(s^2+1)(s+3)}
end{bmatrix}
$$
Of course you could use a different distribution of the pole polynomial and you could throw any two unimodular matrices, which are each others inverse, in-between the two two diagonal matrices. So splitting a transfer function matrix into a minimum phase and non-minimum phase part will in general not be unique.
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
If you define the zeros of a transfer function as the values for $s$ at which it loses rank (also known as transmission zeros) then you could make use of the Smith McMillan form. Namely by multiplying any MIMO transfer function matrix by its pole polynomial one can obtain a polynomial matrix $P(s)$. Any polynomial can be written as
$$
P(s) = U_1(s),D(s),U_2(s)
$$
where $U_1(s)$ and $U_2(s)$ are unimodular (does not lose rank for any $s$) and $D(s)$ is diagonal (does loses rank at $s$ equal to a zero). This diagonal matrix can be split up into two diagonal matrices, one with only right half plane zeros and the other with only left half plane zeros.
For example
$$
G(s) =
begin{bmatrix}
frac{s^4+s+1}{(s^2+1)(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+3)} \
frac{s-1}{(s+1)(s+3)} & frac{1-s}{(s^2+1)(s+1)(s+3)}
end{bmatrix}
$$
can also be written as
begin{align}
G(s) &= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix} \
&= frac{1}{(s^2+1)(s+1)(s+3)}
begin{bmatrix}
1 & s+1 \
0 & 1
end{bmatrix}
begin{bmatrix}
s+2 & 0 \
0 & 1
end{bmatrix}
begin{bmatrix}
1 & 0 \
0 & s-1
end{bmatrix}
begin{bmatrix}
1 & 0 \
s^2+1 & -1
end{bmatrix}
end{align}
Multiplying the left unimodular matrix with the left diagonal matrix and the right diagonal matrix with right unimodular matrix and distribute the terms of the pole polynomial gives
$$
G(s) =
begin{bmatrix}
frac{s+2}{s+1} & 1 \
0 & frac{1}{s+1}
end{bmatrix}
begin{bmatrix}
frac{1}{(s^2+1)(s+3)} & 0 \
frac{s-1}{s+3} & frac{1-s}{(s^2+1)(s+3)}
end{bmatrix}
$$
Of course you could use a different distribution of the pole polynomial and you could throw any two unimodular matrices, which are each others inverse, in-between the two two diagonal matrices. So splitting a transfer function matrix into a minimum phase and non-minimum phase part will in general not be unique.
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
answered Jan 18 at 16:30
Kwin van der VeenKwin van der Veen
5,4952828
5,4952828
add a comment |
add a comment |
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$begingroup$
There is no unique way of doing this, it will only tell you how to split the zeros, but not the poles.
$endgroup$
– Kwin van der Veen
Jan 18 at 12:09
$begingroup$
If $m=1$, then it is easy to decompose by simply finding the unstable zeros. However, for MIMO system, even if I can find unstable zeros for each $g_{ij}(s)$, it's still hard to obtain the final multiplication form.
$endgroup$
– user4756766
Jan 18 at 12:14
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So you do not care if one of the final transfer functions is not causal, otherwise you still also have to choose how to split up the poles, even for $m=1$.
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– Kwin van der Veen
Jan 18 at 12:16
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What if I do not care about whether it's causal. Is there any simple method to realize the decomposition? Could you give me some hint about that @KwinvanderVeen
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– user4756766
Jan 18 at 12:22